$\newcommand{\test}{The Wikibook about \LaTeX} \newcommand{\manugood}{{}_š} \newcommand{\agrigood}{{}_š½}$ Can't specialize in manufactured goods. $š = z_{\manugood} N_{\manugood} + \gamma z_{\manugood} + z_{\manugood} \gamma + z_š \gamma +$

# Autarky

### Assumptions:

• Two goods: manufactured goods and $$š$$ agricultural goods $$š½$$
• Consumer needs to have non-zero food, but can have zero manufactured goods and still survive.
• $U=\ln (š½) + \gamma \ln (š+1)$
• Both goods are produced with CRS technology.
• $Y_š = z_š N_š$ $Y_š½ = z_š½ N_š½$
• No leisure decisions, I guess, just to keep things easy. Set price of š½ as numeraire?

### Formal Definition of Equilibrium:

Click to expand definition
###### Consumer's Problem:
Taking prices $$\{w,p\}$$ as given, chooses $$\{š,š½\}$$ to solve $\max_{š,š½} \ln (š½) + \gamma \ln (š+1)$ subject to the constraints that $š,š½ \geq 0$ $pš + š½ \leq w + \pi_š + \pi_š½$
###### Manufacturing Firm's Problem:
Taking prices $$\{w,p\}$$ as given, chooses $$\{N_š\}$$ to solve $\max p z_š N_š - w N_š$
###### Agricultural Firm's Problem:
Taking prices $$\{w,p\}$$ as given, chooses $$\{N_š½\}$$ to solve $\max z_š½ N_š½ - w N_š½$
###### Market Clearing:
$N_š + N_š½ = 1$ $š = z_š N_š$ $š½ = z_š½ N_š½$

### Solution to firms' problems:

For non-zero finite output, it must be that $p z_š = w = z_š½$ and for zero output of š, it must be that $w \geq p z_š$ In either case, profits will be zero.

### Solution to consumer's problem:

In an interior solution where $$š>0$$, $\frac{(š+1)}{\gamma š½} = p$ $pš + š½ = w$ $š½ = \frac{w+p}{\left(1+\gamma\right)}$ $š = \frac{w\gamma-p}{p\left(1+\gamma\right)}$ And in a solution where $$š=0$$, $š½ = w$ Note that interior solution requires that $$p < w\gamma$$

Click to expand work for consumer problem Assume profits are zero. Will be true in equilibrium, so no big deal. Then Assuming interior solution where $$š½,š>0$$, Lagrangian is: $\mathcal{L} = \ln (š½) + \gamma \ln (š+1) - \lambda [ pš + š½ - w ]$ FOCs: ${1 \over š½} = \lambda$ ${\gamma \over (š+1)} = \lambda p$ $pš + š½ = w$ Implies ${\gamma \over (š+1)} = {1 \over š½} p$ Solve this to express each variable in terms of the other: $š½=š\frac{p}{\gamma}+\frac{p}{\gamma}$ $š=š½\frac{\gamma}{p}-1$ Combine these with budget constraint to solve explicitly for allocation variables: $š½ = \frac{w+p}{\left(1+\gamma\right)}$ $š = \frac{w\gamma-p}{p\left(1+\gamma\right)}$

### Combine problems to get equilibrium

If $$\gamma z_š > 1$$, then we have the interior solution: $š½ = z_š½ \cdot \left( \frac{1+z_š}{z_š\left(1+\gamma\right)} \right)$ $š = \frac{\gamma z_š - 1}{\left(1+\gamma\right)}$ $N_š½ = \frac{š½}{z_{š½}} = \frac{\left(z_{š}+1\right)}{z_{š}\left(1+\gamma\right)}$ $N_{š} = \frac{š}{z_{š}}=\frac{\gamma-\frac{1}{z_{š}}}{1+\gamma}$ Otherwise, if $$\gamma z_š \leq 1$$, $N_š½ = 1$ $N_š = 0$ $š½ = w = z_š½$

Click to expand work for solving for the equilibrium allocations For an interior solution, we have the following characterizing equations:
From consumer: $\frac{(š+1)}{\gamma š½} = p$ $pš + š½ = w$ $š½ = \frac{w+p}{\left(1+\gamma\right)}$ $š = \frac{w\gamma-p}{p\left(1+\gamma\right)}$ $š½,š>0$ From firms: $p z_š = w = z_š½$ From Market Clearing $N_š + N_š½ = 1$ $Y_š = z_š N_š$ $Y_š½ = z_š½ N_š½$
Note how this implies that $p=\frac{ z_š½ }{ z_š }$ And substitute out the prices to get allocations: $š½ = \frac{ z_š½ + {z_š½ \over z_š} }{\left(1+\gamma\right)} = z_š½ \frac{1+z_š}{z_š\left(1+\gamma\right)}$ $š = \frac{z_š½ \gamma-{z_š½ \over z_š}}{{z_š½ \over z_š}\left(1+\gamma\right)} = \frac{\gamma z_š - 1}{\left(1+\gamma\right)}$ $N_š½ = \frac{š½}{z_{š½}} = \frac{\left(z_{š}+1\right)}{z_{š}\left(1+\gamma\right)}$ $N_{š} = \frac{š}{z_{š}}=\frac{\gamma-\frac{1}{z_{š}}}{1+\gamma}$ Note that the above equations are consistent with $$š>0$$ iff $$\gamma z_š > 1$$. If this isn't the case, then it must instead be that $$š=0$$ and $N_š½ = 1$ $N_š = 0$ $š½ = w = z_š½$